我正在尝试为未知数量的数组获得所有可能的组合：
这是JSON数据结构
[
{
"department": "CIS",
"name": "Intro to CIS",
"sections": [
{
"sectionNumber": "01",
"regNum": "012345",
"days": "MWF",
"startTime": "900",
"endTime": "1030",
"labDay": "M",
"labStartTime": "1300",
"labEndTime": "1500"
},
{
"sectionNumber": "02",
"regNum": "098304",
"days": "TR",
"startTime": "1300",
"endTime": "1500",
"labDay": "",
"labStartTime": "",
"labEndTime": ""
}
]
},
{
"department": "MATH",
"name": "Intro to MATH",
"sections": [
{
"sectionNumber": "01",
"regNum": "012345",
"days": "MWF",
"startTime": "900",
"endTime": "1030",
"labDay": "M",
"labStartTime": "1300",
"labEndTime": "1500"
},
{
"sectionNumber": "02",
"regNum": "098304",
"days": "TR",
"startTime": "1300",
"endTime": "1500",
"labDay": "",
"labStartTime": "",
"labEndTime": ""
}
]
}
]
我希望结果如下：结果应包含整个课程关联数组，以便我可以访问其所有数据（sectionNumber，regNum，days，.. etc） < / p>
注意，我不知道会有多少课程或部分。此外，一门课程永远不应该与自己比较。
所需结果：
[
[ [CIS 01 Array],[MATH 01 Array] ],
[ [CIS 01 Array],[MATH 02 Array] ],
[ [CIS 02 Array],[MATH 01 Array] ],
[ [CIS 02 Array],[MATH 02 Array]]
]
每个结果数组应包含数组的数组（与课程数相同）。
我想我应该使用笛卡尔积，但我不完全确定如何实现它。 我开始做这样的事情，但我知道这不正确：
for($i = 0; $i < count($json_data); $i++){ //courses
for($k = 0; $k < count($json_data[$i]["sections"]); $k++){ //first sections
for($p = 1; $p < count($json_data[$i]["sections"]); $p++){ //all other courses but 1
for($h = 0; $h < count($json_data[$i]["sections"][$p]); $h++){
echo $json_data[$i]["sections"][$k]["labDay"];
}
}
}
}
答案 0 :(得分：0)
答案 1 :(得分：0)
这是我的解决方案：
//simplified json for answer, you could use yours
$json = '[
{
"department": "CIS",
"sections": [
{
"sectionNumber": "CIS-1"
},
{
"sectionNumber": "CIS-2"
}
]
},{
"department": "ENG",
"sections": [
{
"sectionNumber": "ENG-1"
},
{
"sectionNumber": "ENG-2"
}
]
},
{
"department": "MATH",
"sections": [
{
"sectionNumber": "MATH-1"
},
{
"sectionNumber": "MATH-2"
}
]
}
]';
$data = json_decode($json, true);
$cartesian = [];
//Use array map to run over all array items
array_map(function($item) use($data, &$cartesian) {
//starting from your element, search for all others "next departments"
for($i = array_search($item, $data)+1, $c = count($data); $i<$c; $i++) {
//foreach "next departments" get section
foreach($data[$i]['sections'] as $section) {
//foreach sections of current department, do
foreach($item['sections'] as $item_section) {
//append to cartesian resultset
$cartesian[] = [$item_section, $section];
}
}
}
}, $data);
//create a reverse array, to get all reverse combinations.
// Ex.: We have CIS-1 -> MATH-1, now we have MATH-1 -> CIS-1
$reverse = array_map('array_reverse', $cartesian);
//merge to cartesian resultset
$cartesian = array_merge($cartesian, $reverse);
print_r(count($cartesian)); print_r($cartesian);
输出：
24
Array
(
[0] => Array
(
[0] => Array
(
[sectionNumber] => CIS-1
)
[1] => Array
(
[sectionNumber] => ENG-1
)
)
[1] => Array
(
[0] => Array
(
[sectionNumber] => CIS-2
)
[1] => Array
(
[sectionNumber] => ENG-1
)
)
[2] => Array
(
[0] => Array
(
[sectionNumber] => CIS-1
)
[1] => Array
(
[sectionNumber] => ENG-2
)
)
[3] => Array
(
[0] => Array
(
[sectionNumber] => CIS-2
)
[1] => Array
(
[sectionNumber] => ENG-2
)
)
[4] => Array
(
[0] => Array
(
[sectionNumber] => CIS-1
)
[1] => Array
(
[sectionNumber] => MATH-1
)
)
[5] => Array
(
[0] => Array
(
[sectionNumber] => CIS-2
)
[1] => Array
(
[sectionNumber] => MATH-1
)
)
[6] => Array
(
[0] => Array
(
[sectionNumber] => CIS-1
)
[1] => Array
(
[sectionNumber] => MATH-2
)
)
[7] => Array
(
[0] => Array
(
[sectionNumber] => CIS-2
)
[1] => Array
(
[sectionNumber] => MATH-2
)
)
[8] => Array
(
[0] => Array
(
[sectionNumber] => ENG-1
)
[1] => Array
(
[sectionNumber] => MATH-1
)
)
[9] => Array
(
[0] => Array
(
[sectionNumber] => ENG-2
)
[1] => Array
(
[sectionNumber] => MATH-1
)
)
[10] => Array
(
[0] => Array
(
[sectionNumber] => ENG-1
)
[1] => Array
(
[sectionNumber] => MATH-2
)
)
[11] => Array
(
[0] => Array
(
[sectionNumber] => ENG-2
)
[1] => Array
(
[sectionNumber] => MATH-2
)
)
[12] => Array
(
[0] => Array
(
[sectionNumber] => ENG-1
)
[1] => Array
(
[sectionNumber] => CIS-1
)
)
[13] => Array
(
[0] => Array
(
[sectionNumber] => ENG-1
)
[1] => Array
(
[sectionNumber] => CIS-2
)
)
[14] => Array
(
[0] => Array
(
[sectionNumber] => ENG-2
)
[1] => Array
(
[sectionNumber] => CIS-1
)
)
[15] => Array
(
[0] => Array
(
[sectionNumber] => ENG-2
)
[1] => Array
(
[sectionNumber] => CIS-2
)
)
[16] => Array
(
[0] => Array
(
[sectionNumber] => MATH-1
)
[1] => Array
(
[sectionNumber] => CIS-1
)
)
[17] => Array
(
[0] => Array
(
[sectionNumber] => MATH-1
)
[1] => Array
(
[sectionNumber] => CIS-2
)
)
[18] => Array
(
[0] => Array
(
[sectionNumber] => MATH-2
)
[1] => Array
(
[sectionNumber] => CIS-1
)
)
[19] => Array
(
[0] => Array
(
[sectionNumber] => MATH-2
)
[1] => Array
(
[sectionNumber] => CIS-2
)
)
[20] => Array
(
[0] => Array
(
[sectionNumber] => MATH-1
)
[1] => Array
(
[sectionNumber] => ENG-1
)
)
[21] => Array
(
[0] => Array
(
[sectionNumber] => MATH-1
)
[1] => Array
(
[sectionNumber] => ENG-2
)
)
[22] => Array
(
[0] => Array
(
[sectionNumber] => MATH-2
)
[1] => Array
(
[sectionNumber] => ENG-1
)
)
[23] => Array
(
[0] => Array
(
[sectionNumber] => MATH-2
)
[1] => Array
(
[sectionNumber] => ENG-2
)
)
)
答案 2 :(得分：0)
这是一个通用的跨产品解决方案（不是我的，我发现它在某个地方，不记得哪里，但有一堆在线可用）：
function crossProduct() {
$_ = func_get_args();
if (count($_) == 0) {
return array(array());
}
$a = array_shift($_);
$c = call_user_func_array('crossProduct', $_);
$r = array();
foreach ($a as $v) {
foreach ($c as $p) {
$r[] = array_merge(array($v), $p);
}
}
return $r;
}
然而，您需要对解决方案进行一些预处理：
$allDepartmentsSections = [];
foreach ($jsonArrayEntry as $entry) {
$sections = [];
foreach ($entry["section"] as $section) {
$sections[] = $section + [ "department" => $entry["department"], "name"=>$entry["name"] ];
}
$allDepartmentsSections[] = $sections;
}
然后你可以这样做：
call_user_func_array('crossProduct', $allDepartmentsSections);